Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void Fastio() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
double pi = acos(-1);
double adotb(complex<double> a, complex<double> b) {
  return (conj(a) * b).real();
}
double acrossb(complex<double> a, complex<double> b) {
  return (conj(a) * b).imag();
}
double sinseta(complex<double> a, complex<double> b) {
  double res = asin(abs(acrossb(a, b) / (abs(a) * abs(b)))) * 180.0 / pi;
  if (res > 180.0) res -= 180.0;
  return res;
}
long long fact[1000006], inv[1000006];
long long fp(long long base, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base;
  long long ans = fp(base, exp / 2);
  ans = (ans * ans) % 1000000007;
  if (exp % 2 != 0) ans = (ans * (base % 1000000007)) % 1000000007;
  return ans;
}
void calcFacAndInv(long long n) {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % 1000000007;
    inv[i] = fp(fact[i], 1000000007 - 2);
  }
}
long long ncr(long long n, long long r) {
  return ((fact[n] * inv[r]) % 1000000007 * inv[n - r]) % 1000000007;
}
long long npr(long long n, long long r) {
  return (fact[n] * inv[n - r]) % 1000000007;
}
int main() {
  Fastio();
  double n, h;
  cin >> n >> h;
  double area = 0.25 * h;
  for (int i = 1; i < n; i++) {
    double res = h * h * ((i / n * area) / area);
    cout << fixed << setprecision(9) << sqrt(res) << " ";
  }
  return 0;
}